Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
#define N 100009
#define M 20000009
using namespace std;

int n,m,cnt,trtot,rt,a[N],b[N],lf[N],rg[N],ls[M],rs[M],tg[M];
ll ans=1ll<<60,f[N];
void mdy(int &k,int x){
	if (!k) k=++trtot; tg[k]=x;
}
void pushdn(int k){
	if (tg[k]){
		mdy(ls[k],tg[k]); mdy(rs[k],tg[k]);
		tg[k]=0;
	}
}
void mdy(int &k,int l,int r,int x,int y,int z){
	if (!k) k=++trtot;
	if (x<=l && r<=y){
		mdy(k,z); return;
	}
	int mid=l+r>>1;
	pushdn(k);
	if (x<=mid) mdy(ls[k],l,mid,x,y,z);
	if (y>mid) mdy(rs[k],mid+1,r,x,y,z);
}
int qry(int k,int l,int r,int x){
	if (tg[k] || l==r) return tg[k];
	int mid=l+r>>1;
	pushdn(k);
	return x<=mid?qry(ls[k],l,mid,x):qry(rs[k],mid+1,r,x); 
}
int calc(int x,int y,int z){
	if (x<=z && z<=y) return 0;
	return (x-z+m)%m;
}
void solve(int k,int l,int r){
	if (tg[k]){
		ans=min(ans,f[tg[k]]+(tg[k]>cnt?0:calc(l,r,rg[tg[k]]))); return;
	}
	if (!k || l==r) return;
	int mid=l+r>>1;
	pushdn(k);
	solve(ls[k],l,mid); solve(rs[k],mid+1,r);
}
int main(){
	scanf("%d%d",&n,&m);
	int i,x,y; ll res=0,tx=0;
	for (i=1; i<=n; i++){
		scanf("%d%d",&x,&y); x<<=1;
		res+=x; tx+=x;
		if (y==1){
			if (x>m){
				puts("-1"); return 0;
			}
			if (!x) continue;
			a[++cnt]=tx%m; b[cnt]=x%m; tx=0;
		}
	}
	for (i=1,tx=0; i<=cnt; i++){
		lf[i]=(a[i]+tx)%m; rg[i]=(a[i]-b[i]+m+tx)%m;
		tx=(tx+a[i])%m;
	}
	mdy(rt,cnt+1);
	for (i=cnt; i; i--){
		x=qry(rt,0,m-1,rg[i]);
		f[i]=(x>cnt?0:f[x]+(rg[i]-rg[x]+m)%m);
		if (lf[i]<=rg[i]){
			if (lf[i]) mdy(rt,0,m-1,0,lf[i]-1,i);
			if (rg[i]<m-1) mdy(rt,0,m-1,rg[i]+1,m-1,i);
		} else
			mdy(rt,0,m-1,rg[i]+1,lf[i]-1,i);
	}
	solve(rt,0,m-1);
	printf("%lld\n",res+ans);
	return 0;
}